IID Suppose X1,..., X, N(u,o2). Construct a 95% confidence interval for o? when µu is known.
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Q: Give a 99.5% confidence interval, for μ₁ 2.63, S1 1 3.11, s2 0.6 n1 40, x₁ 20, ₂ = n₂ = है H - - = 2…
A: given datan1=40n2=20x1=2.63x2=3.11s1=1s2=0.6CI=0.9950
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- The patient's recovery time from a particular surgical procedure is normally distributed with a mean of 26 days and a standard deviation of 7.24 days. Let X - is the number of days a randomly selected patient needs to recover from the surgical procedure. What is the upper bound of 90% confidence interval of X?What critical value of t* should be used for a 95% confidence interval for the population mean based on a random sample of 21 observations? Find the t-table here. t* = 1.721 t* = 1.725 t* = 2.080 t* = 2.086The average American gets a haircut every 43 days. Is the average larger for college students? The data below shows the results of a survey of 11 college students asking them how many days elapse between haircuts. Assume that the distribution of the population is normal. 38, 37, 38, 45, 53, 55, 35, 37, 35, 41, 38 What can be concluded at the the α = 0.05 level of significance level of significance? For this study, we should use? Select an answer: t-test for a population mean? z-test for a population mean? The null and alternative hypotheses would be: H0:H0? p or μ Select an answer > = ≠ ≤ < ≥ H1:H1? μ or p Select an answer ≠ ≥ > ≤ = < 2. The test statistic ? t z = (please show your answer to 3 decimal places.) The p-value = _______ (Please show your answer to 3 decimal places.) The p-value is? > or ≤ α 3. Based on this, we should?? Select an answer reject ? accept ? or fail to reject ?…
- A random sample of n measurements was selected from a population with unknown mean μ and standard deviation σ=15 for each of the situations in parts a through d. Calculate a 99% confidence interval for μ for each of these situations d. n=110, x=5.97 e. Is the assumption that the underlying population of measurements is normally distributed necessary to ensure the validity of the confidence intervals in parts a through d? Explain.A sample of 66 night-school students' ages is obtained in order to estimate the mean age of night-school students. x = 24.7 years. The population variance is 18. (a) Give a point estimate for u. (Give your answer correct to one decimal place.) (b) Find the 95% confidence interval for u. (Give your answer correct to two decimal places.) Lower Limit Upper Limit (c) Find the 99% confidence interval for u. (Give your answer correct to two decimal places.) Lower Limit Upper Limita sample size n=99 is drawn from a population whose standard deaviation is 11 (a) find the margin of error for a confidence interval for u (population mean). round the answer to at least three decimal places the margin of error for a 90% confidence for u (population mean) is
- erproduction of uric acid in the body can be an Indication of cell breakdown. This may be an advance indication of illness such as go adult male patlent has taken eleven blood tests for uric acid. The mean concentration was x = 5.30 mg/dl. The distribution of uric rmal, with o = 1.95 mg/dI. a) Find a 95% confidence interval for the population mean concentration of uric acid in this patient's blood. What is the margin of ei lower limit upper limit margin of error (b) What conditions are necessary for your calculations? (Select all that apply.) normal distribution of uric acid uniform distribution of uric acid o is known On is large Oo is unknown (c) Interpret your results in the context of this problem. O The probability that this interval contains the true average uric acid level for this patient is 0.95. O We are 5% confident that the true uric acid level for this patient falls within this interval. O We are 95% confident that the true uric acid level for this patient falls within…“Domestic cats kill many more wild birds in the United States than scientists thought”, states a recent article. Researchers used a random sample of ? = 300 households in the US with cats to estimate that 25% of household cats in the US hunt outdoors Using the normal distribution to find a 90% confidence interval for the population proportion ?. (Take z* = 1.645) a. Can we assume that the sample size is large enough to use the standard normal distribution? Why or why not? b. What is the best point estimate for ?? c. What is the margin of error without using StatKey? d. Find and interpret a 90% confidence interval for the proportion of household cats in the US that hunt outdoors.ou intend to estimate a population mean with a confidence interval. You believe the population to have a normal distribution. Your sample size is 6.a.) The degrees of freedom df=df=b.) Use the t-table on page 4 to find the critical value tctc that corresponds to a confidence level of 95%.tc=(Use 4 decimal places)
- A sample of size n=79 Is drawn from a population whose standard deviation is o = 29, Part 1 of 2 (a) Find the margin of error for a 90% confidence interval for l. Round the answer to at least three decimal places. The margin of error for a 90% confidence interval for u is Part 2 of 2 (b) If the sample size weren=44, would the margin of error be larger or smaller? (Choose one) ▼ because the sample size is (Choose one) Larger SmallerThe marginal error for confidence intervals concerning μ doesn't depend on the sample size. However, it depends on the critical value t*, as well as the standard deviation of the observed sample. Is this true or false? And why?
